1. How much force does it take to move a 500kg box with an acceleration of 2 m/s/s?

It states that an object stays in rest or in uniform motion unless and until compelled by an external unbalanced force . This is also known as " Law of Inertia " .

Ex - (i) We tend to fall backwards while standing in the bus , when it starts suddenly.

(ii) We tend to fall forward while standing in the bus , when it stops suddenly.

Newton's Second Law of Motion :-

It states that the rate of change of momentum is directly proportional to the applied force in the direction of the force . That is ∆p $\propto$ Force .

Ex - (i) When we push car & a truck with same force the car will have Greater acceleration due to less mass . (Imaginary case ).

(ii) We feel easy to push a empty cart than a full cart .

Newton's Third Law of Motion :-

It states that every action has equal and opposite reaction .

Ex - (i) We are able to walk on ground due to Newton's third law of motion .

(ii) When we push wall it also applies equal and opposite force .

8 0

3 years ago

Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre

Anni [7]

Answer:

a) k = Mg / d , b) v = √2gh , c) v_{f} = $\frac{2}{3} \ \sqrt{2gh}$ , d) x² + 6d x - $\frac{8}{3}$ dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

∑ F = 0

$F_{e}$ - W = 0

k d = Mg

k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

Em₀ = U = m g h

lowest point. Right at the point of shock

$Em_{f}$ = K = ½ m v²2

as there is no friction, energy is conserved

Em₀ = Em_{f}

mg h = ½ m v²

v = √2gh

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

p_{f} = (2M + M) v_{f}

the moment is preserved

p₀ = p_{f}

2M v = 3M v_{f}

v_{f} = ⅔ v

v_{f} = $\frac{2}{3} \ \sqrt{2gh}$

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

Em₀ = K = ½ (3M) $v_{f}^2$

Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

Em₀ = 4/3 M gh

final point. With the spring fully compressed

Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

Em₀ = Em_f

4/3 M g h = ½ k x² + 3M g x

½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

$\frac{1}{2}$ ( $\frac{Mg}{d}$ ) x² + 3Mg x - $\frac{4}{3}$ Mgh = 0

$\frac{x^{2} }{2d}$ + 3 x - $\frac{4}{3}$ h = 0

to find the value of the spring compression, the second degree equation must be solved

x² + 6d x - $\frac{8}{3}$ dh = 0

x = [-6d ± $\sqrt{(36 d^{2} - 4 \frac{8}{3} dh) }$ ] / 2

x = [-6d ± 6d $\sqrt{ 1 - \frac{32}{3 \ 36} \ \frac{h}{d} }$ ]/2

x = 3d ( -1± $\sqrt{ 1 - 0.296 \frac{h}{d} }$ )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0

3 years ago